Hi,

I need to find the inverse Laplace transform of

$\displaystyle F(s) = s^{-2} \tanh(sT/2)$.

I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips?

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- Nov 6th 2009, 05:24 PMharbottleInverse Laplace Transform
Hi,

I need to find the inverse Laplace transform of

$\displaystyle F(s) = s^{-2} \tanh(sT/2)$.

I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips? - Nov 6th 2009, 05:33 PMmr fantastic
- Nov 6th 2009, 06:11 PMharbottle
Sorry, typo'd :(. Was thinking of periods while I was writing it down, and an n found its way in there somewhere

- Nov 7th 2009, 01:59 AMmr fantastic
- Nov 7th 2009, 02:35 AMharbottle
I haven't used the residue theorem for a while, I know. The Bromwich integral I know by definition, but I don't think we're expected to use it (and haven't been taught how).

Anyway, I'll be keen to try whatever you're thinking, or if you know a way of doing it without these things, that's fine too. - Nov 7th 2009, 02:39 AMmr fantastic
- Nov 7th 2009, 06:19 PMchisigma
You are 'lucky' because in this case You don't need to know sophisticated tools like Bromwitch integral, residue theorem and so on but only to have 'little imagination' (Wink) ...

The LT to be inverted is...

$\displaystyle F(s)= \frac{1}{s^{2}}\cdot \tanh \frac{sT}{2}$ (1)

Now, taking into account that...

a) for $\displaystyle |s|>0$ the function $\displaystyle \tanh (*)$ can be written as...

$\displaystyle \tanh \frac{sT}{2} = \frac{1 - e^{-sT}}{1+e^{-sT}} = (1-e^{-sT})\cdot (1 - e^{-sT} + e^{-2sT} - \dots)$ (2)

b) is...

$\displaystyle \mathcal{L}^{-1} \{\frac{1}{s^{2}}\cdot (1-e^{-sT}) \}=t \cdot H(t) -(t-T)\cdot H(t-T)$ (3)

... that is the function represented here...

http://digilander.libero.it/luposabatini/MHF27.bmp

... is easy enough verify with the aid of (2) that the inverse LT we are searching is the function (3) 'periodized with alternating signs' , i.e. the periodic function represented here...

http://digilander.libero.it/luposabatini/MHF28.bmp

So your 'intuition' was exact: we have a periodic function (Clapping)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 7th 2009, 07:12 PMmr fantastic
- Nov 8th 2009, 06:02 AMshawsend
Hi. Just for fun:

$\displaystyle \mathcal{L}^{-1}\left\{\frac{1}{s^2}\tanh(sT/2)\right\}=\frac{1}{2\pi i}\mathop\int\limits_{c-i\infty}^{c+i\infty}e^{st}\frac{1}{s^2}\tanh(sT/2)ds$

and with some effort, we can show a closed contour pinned at the line $\displaystyle c+i\gamma$ and allowed to expand without bounds, reduces to the Bromwich integral and sum of the enclosed residues:

$\displaystyle \mathcal{L}^{-1}\left\{\frac{1}{s^2}\tanh(sT/2)\right\}=\mathop\text{Res}\limits_{z=0}\left\{e^ {st}\frac{1}{s^2}\tanh(sT/2)\right\}$$\displaystyle +\sum_{n=0}^{\infty}\mathop\text{Res}\limits_{z=z_ n}\left\{e^{st}\frac{1}{s^2}\tanh(sT/2)\right\},\quad z_n=\pm\frac{(2n+1)\pi i}{T}$

$\displaystyle =\frac{T}{2}-\frac{4T}{\pi^2}\sum_{n=0}^{\infty}\frac{1}{(2n+1) ^2}\cos\left[\frac{(2n+1)\pi x}{T}\right]$

and I suspect that is the Fourier series of the answer obtained by Chisigma and suggested by Mr. F. - Nov 8th 2009, 08:01 AMCaptainBlack
- Nov 10th 2009, 12:06 AMharbottle
Nice! That looks easy now with $\displaystyle \chi\sigma$'s hint.

And I'll have a look at your solution too shawsend though as Mr.F suggested I might have some difficulties.

Thanks everyone!